TA的每日心情 | 擦汗 2016-4-17 12:18 |
|---|
签到天数: 1 天 [LV.1]炼气
|
本帖最后由 tingsanguo 于 2014-5-2 14:01 编辑 / S/ N) ~ _+ ?4 g E
水风 发表于 2014-5-2 01:57 ; r8 \" G: d, n) j. E
再次重温自己与数学天才们之间的差距有多大了。解出来了,我还看不懂思路 ... ; z4 a9 |7 {2 B0 j$ X# D
9 r" X& l- K. J, V
fixed point f(X) = X = (1-n), where n is the number of people who divide the coconut.
1 b3 s1 V1 c+ f8 B$ x7 l( b% d& L, M/ F' T0 u
The least number of coconut before dividing is Z = [n^(n+1) + X], i.e., always divide (n+1) times.
1 v' x( h4 W. f+ C% ~3 J
1 o3 k6 u+ q- H. i4 R2 x" d r& pAfter being divided (n+1) times, Z becomes 1, since X is fixed point.: r( y& G1 o9 f
& b1 x% b1 }' }, v; A: Mn n^(n+1)+X, n^(n)+X, n^(n-1)+X, n^(n-2)+X ! c$ j. \" |- [/ [ O0 F" v
1 1
2 p, r/ m0 ]$ u5 }. z2 7 3 1 3 \ U5 ?# S( U2 S
3 79 25 7 1
* j# C, a" F" A- ?- E0 D- N E4 1021 253 61 13 1 . H7 g$ m6 \8 @
5 15621 3121 621 121 21 1
5 h. c4 u* r, N+ X7 V& ^+ C% T5 O; U, a) P4 n) {
. }' v5 S! M$ }. @. _+ r |
|